Mcleod Motor WAC

September 4, 2016 | Author: Abdul Khan | Category: Types, School Work
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Mcleod Motors Written analysis. expain the operations and calculations of the case...

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Technology Operations & Management Case: Mcleod Motors Ltd. WAC Submission

Prepared by: Abdul Samad Khan (08975) Submitted to: Mr. Kamran Mumtaz Date: 23rd October 2014

Executive Summary McLeod Motors LTD is a successful company with numerous customers in the original equipment manufacturer (OEM) market. These customers use motors as components in large products and in the replacement market. They have made over 40 models of electronic motors. During the previous year McLeod engineers have recommended that the company minimize the different number of end shields and replace these different end shields with a new end shield called BN-88-55. The new end shield replaces 15 end shields in one motor product line of motors. Sue Reynolds, plant manager at McLeod Motors LTD factory realized that this new product would result in lower manufacturing costs, make higher sale and lower inventories; however, the results that the vice president expected to see at the end of the period did not occur. McLeod Motors LTD has too much inventory and the warehouse does not have enough space for all their inventories such as: Raw Materials, Work in process, finish goods and supplies. Sue Reynolds needs to investigate the reasons why the warehouse does not have enough space and what Macleod has to do in order to decrease their inventory and lower the cost of keeping the inventory in the warehouse. Sue Reynolds has to consider what action to take and what should tell John Ingram. Process Flow

WIP

Drilling

WIP

Milling

WIP

Turning

FP

Assembly

Above flow diagram represents the process flow at Mcloed, where there is a Work-in-process inventory after every process except for assembly. One the product has passed from the assembly, its stored in the inventory (ware house) as a finished good.

Operations

Departm ent

Machine

Tap 4 holes, concave face Tap 4 holes, convex face Turn convex face Turn concave face Inspect and Finish

Drilling

 

Drill press

Pieces per machine hour 75

Pieces per machine per minute 1.25

Machines available in department 9

Drilling

Drill press

75

1.25

9

Turning

Lathe

100

1.67

5

Turning

Lathe

120

2.0

5

Inspection

Work bench and hand tools

150

2.5

3

Through put time: 8.67 min Cycle Time: 2.5 min

Analysis: 

  

Since the company has decided to reduce the number of product from a variety of 36 SKUs to 5 types, Mcloed is going towards Standardization to increase the volumes and reduce the cost and customization. All the machines are assembled in way to facilitate the process, so it is a processed based – Batch process Working hours = 8 hours x 5 Days = 40 hours per week Supply market: o OEM (Original equipment manufacturer) was the main customer with known annual forecast o Secondary customer was replacement market with a more fluctuating short time forecast

Flexibility

Quality

Speed

Cost

Calculation:    



Annual demand = 2500 (weekly) x 52 (weeks) = 130,000 units Actual cost per unit = 5.28 (direct cost) + 0.77 (direct Labor) + 2.31 (Overhead cost) = $8.36 per unit Selling price= 8.36 + 20% profit = $10 Weekly demands: o Average weekly demand= 165 (units) x 15 (shields) = 2475 units per week o Actual weekly demand = 2500 units per week o Standard weekly output = 1248 (units) x 2 (batches) = 2496 units per week Direct Labor time on one BN-88-55s = 3.1 min

Standard Demand and Inventory Analysis:  Order Size (Q) = 2500 per week  Unit Cost (C) =$8.3  Cost of Ordering (S) = 10 min [3.1 min = $0.77] + unit cost= $7.45 + $8.3 =$15.75  Annual Demand (R) = 2500 * 52 = 130,000 units  Annual inventory carrying costs (K) =? Q=(2RS/KC) 2500=(2*130000*15.75)/(K*8.3) K=7.86% (annual inventory carrying cost as a fraction of unit cost)  Annual Inventory Cost = KCQ/2 = (.0786*8.3*2500)/2 = $815 Standard annual Inventory (Q/2)*52 = 1250*52= 65000 units per year Actual Demand and Inventory Analysis (Before BN-88-55):  Order Size (Q) = 15 * 165 units = 2475 units per week  Unit Cost (C) =$8.3  Cost of Ordering (S) = 10 min [3.1 min = $0.77] + unit cost= $7.45 + $8.3 =$15.75  Annual Demand (R) = 2475 * (52+2) =133,650 units  Annual inventory carrying costs (K) =?

Q=(2RS/KC) 2475=(2*133650*15.75)/(K*8.3) K=8.28% (annual inventory carrying cost as a fraction of unit cost) 

Annual Inventory Cost = KCQ/2 = (.0828*8.3*2475)/2 = $850.5 Additional inventory cost = $850.5 + 25% = 815 + 212.6 = $1063.1 Standard annual Inventory (Q/2)*54 = 1237.5*54= 66,825 units per year Actual Demand and Inventory Analysis (After BN-88-55):  Order Size (Q) = 2496  Unit Cost (C) =$8.3  Cost of Ordering (S) = 10 min [3.1 min = $0.77] + unit cost= $7.45 + $8.3 =$15.75  Annual Demand (R) = 2496*54= 134,784 units  Annual inventory carrying costs (K) =? Q=(2RS/KC) 2496=(2*134784*15.75)/(K*8.3) K=8.21% (annual inventory carrying cost as a fraction of unit cost)  Annual Inventory Cost = KCQ/2 = (.0821*8.3*2500)/2 = $850.5 Additional inventory cost = $815 + 25% = 815 + 212.6 = $1063.1 Standard annual Inventory (Q/2)*54 = 1248*54= 67,392 units per year

Conclusion: It can been seen through the calculation that before BN-88-55, the batch size was of 2475 units per week, which was giving us an annual inventory of 66,825 units per year. When we increased our batch size to 2496 according to new process, our annual inventory

increased to 67,392 units. There is an increase of 567 units in the inventory, which adds to additional 25% of inventory cost. We can see that our actual required inventory should be 65000 units, but instead we are producing a surplus of 2392 units by keep a buffer of two week. Sue can suggest a decrease in the buffer by a week. This will allow them a reduction in inventory by 1248 unit. The new inventory number will be 66,144 units. Through this method then can also meet the immediate demand of the replacement market segment as well.

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